Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{10}{x + 4} \times \dfrac{8x^2 + 32x}{8x} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 10 \times (8x^2 + 32x) } { (x + 4) \times 8x } $ $ a = \dfrac {10 \times 8x(x + 4)} {8x (x + 4)} $ $ a = \dfrac{80x(x + 4)}{8x(x + 4)} $ We can cancel the $x + 4$ so long as $x + 4 \neq 0$ Therefore $x \neq -4$ $a = \dfrac{80x \cancel{(x + 4})}{8x \cancel{(x + 4)}} = \dfrac{80x}{8x} = 10 $